how much of 0.1 M NaOH is needed to neutralize 5mL of 0.1 M of HCL to a ph 7(color on the ph scale)...Thank you
mL1 x M1 = mL2 x M2
Substitute and solve for the unknown.
mL*0.1M = 5 mL x 0.1M
mL = ?
To determine how much of 0.1 M NaOH is needed to neutralize 5 mL of 0.1 M HCl to a pH of 7, we first need to understand the concept of neutralization and how to calculate it.
Neutralization is a chemical reaction between an acid and a base that results in the formation of a salt and water. In this case, the reaction is between HCl (the acid) and NaOH (the base), resulting in the formation of NaCl (salt) and water (H2O).
To calculate the amount of NaOH needed, we can use the principle of stoichiometry and the formula:
Ma * Va = Mb * Vb
where Ma is the molarity of acid (HCl), Va is the volume of acid, Mb is the molarity of base (NaOH), and Vb is the volume of base.
Given:
Molarity of HCl (acid) = 0.1 M
Volume of HCl (acid) = 5 mL
Molarity of NaOH (base) = 0.1 M
pH to be achieved = 7
We know that the pH of 7 corresponds to a neutral solution. At pH 7, the concentration of H+ (hydronium ions) and OH- (hydroxide ions) is equal, which means we have a balanced solution between acid and base.
Given the molarities of both the acid and the base are the same (0.1 M), we can assume that the ratio between the acid and base is 1:1. Therefore, to neutralize 5 mL of 0.1 M HCl, we need an equal amount of 0.1 M NaOH.
Hence, the amount of 0.1 M NaOH needed to neutralize 5 mL of 0.1 M HCl to a pH of 7 is also 5 mL.